6 Exercise 3: Suppose RO = 0x20000000 and R1 = 0x12345678, find the result storedin memory for the following operations. State the contents of the memory and theiraddress locations in hexadecimal after the following operations are executed. Drawa table (similar to the previous question) to show the memory value if the processoruses Little Endian.a) STR R1, [RO]b) STR R1, [R0, #4]Exercise 4: Suppose RO = 0XOFOFOFOF and R1 = 0x12345678, find the result in R3 ofthe following operations. State the contents of the register in hexadecimal.%3D%3Da) ADD R3, R1, ROb) SUB R3, R1, ROc) AND R3, R1, ROd) ORR R3, R1, ROe) EOR R3, R1, RO

exercise 3 a)STR R1, [R0] The guidance stores the worth of r1 to memory at the location that the…

Answered: Exercise 3: Suppose RO = 0x20000000 and…

C++ Programming: From Problem Analysis to Program Design

8th Edition

ISBN:9781337102087

Author:D. S. Malik

Publisher:D. S. Malik

Chapter12: Points, Classes, Virtual Functions And Abstract Classes

Section: Chapter Questions

Problem 19SA

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Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.
Modiflow y the beprogram given to include response time program;ROUND ROBIN CPU SCHEDULING ALGORITHM #include<stdio.h> #include<conio.h> using namespace std; int main() { int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max; float awt=0,att=0,temp=0; printf("Enter the no of processes--"); scanf("%d",&n); for(i=0;i<n;i++) { printf("\nEnter Burst Time for process %d--", i+1); scanf("%d",&bu[i]); ct[i]=bu[i]; } printf("\nEnter the size of time slice--"); scanf("%d",&t); max=bu[0]; for(i=1;i<n;i++) if(max<bu[i]) max=bu[i]; for(j=0;j<(max/t)+1;j++) for(i=0;i<n;i++) if(bu[i]!=0) if(bu[i]<=t) { tat[i]=temp+bu[i];…
Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:
P1:Given the following memory values and a one-address machine with an accumulator,whatvalues do the following instructions load into the accumulator?·Word 20 contains 40.·Word 30 contains 50.·Word 40 contains 60.·Word 50 contains 70.a. LOAD IMMEDIATE 20b. LOAD DIRECT 20c. LOAD INDIRECT 20d. LOAD IMMEDIATE 30e. LOAD DIRECT 30f. LOADINDIRECT 30
Modiflow y the beprogram given to include response time program;SJF CPU SCHEDULING ALGORITHM #include<stdio.h> #include<conio.h> using namespace std; int main() { int p[20], bt[20], wt[20], tat[20], i, k, n, temp; float wtavg, tatavg; printf("\nEnter the number of processes--"); scanf("%d", &n); for(i=0;i<n;i++) { p[i]=i; printf("Enter Burst Time for Process %d--", i); scanf("%d", &bt[i]); } for(i=0;i<n;i++) for(k=i+1;k<n;k++) if(bt[i]>bt[k]) { temp=bt[i]; bt[i]=bt[k]; bt[k]=temp; temp=p[i]; p[i]=p[k];…
2. 2) You are required to write an Assembly Language program segment to perform theoperation Ci = where Ai and Bi represents a set of 50 memory locations each storing avalue such that the A values are stored starting from memory location 100 while the B valuesare stored starting from memory location 200. The results are to be stored starting frommemory location 300.
Modify below program to include response timeProgram:ROUND ROBIN CPU SCHEDULING ALGORITHM: #include<stdio.h>#include<conio.h>using namespace std; int main(){int i,j,n,bu[10],wa[10],tat[10],t,ct[10],max; float awt=0,att=0,temp=0;printf("Enter the no of processes--"); scanf("%d",&n);for(i=0;i<n;i++){printf("\nEnter Burst Time for process %d--", i+1); scanf("%d",&bu[i]);ct[i]=bu[i];}printf("\nEnter the size of time slice--"); scanf("%d",&t);max=bu[0];for(i=1;i<n;i++)if(max<bu[i])max=bu[i];for(j=0;j<(max/t)+1;j++)for(i=0;i<n;i++)if(bu[i]!=0)if(bu[i]<=t){tat[i]=temp+bu[i];temp=temp+bu[i];bu[i]=0;}else{bu[i]=bu[i]-t;temp=temp+t;}for(i=0;i<n;i++){wa[i]=tat[i]-ct[i];att+=tat[i];awt+=wa[i];}printf("\nThe Average Turnaround time is--%f",att/n);printf("\nThe Average Waiting time is--%f ",awt/n);printf("\n\tPROCESS\t BURST TIME \t WAITING TIME\tTURNAROUND TIME\n");for(i=0;i<n;i++)printf("\t%d \t %d \t\t %d \t\t %d…
1. The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111 = Store AC to I/O In these cases, the 12-bit address identifies a particular I/O device. List the steps for every execution for the following program and illustrate using table that explain the process below : a. Load AC from device 5. b. Add contents of memory location 940. c. Store AC to device 6. d. Assume that the next value retrieved from device 5 is 3 and that location 940 contains a value of 2. Please pointing a, b,c ans. Because one I already upload this question and I didn't understand which one is and of a...please write ans a, b , c please
In each of the following cases, write a short program in Vole to perform the requestedactivities. Assume that each of your programs is placed in memory starting at address 0x00.a. Move the value at memory location 0xD8 to memory location 0xB3.b. Interchange the values stored at memory locations 0xD8 and 0xB3.c. If the value stored in memory location 0x44 is 0x00, then place the value 0x01 in memorylocation 0x46; otherwise, put the value 0xFF in memory location 0x46
Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary:00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101. If the Cache of this computer system is re-organized to a 16-Byte direct mapped cache with a block size of 4Bytes, what would be the hit rate after the 20th memory access is finished?
Assume a computer system has a main memory of 256 Bytes. The following is a memory byte-access trace history of a program run on this system. For example, as it is shown, the program first accesses memory address 0000 0000, and then it accesses memory address 0000 0001 and so on. Note the memory addresses are represented in binary: 00000000, 00000001, 00000010, 00000011, 00001000, 00010000, 00010001, 00000100, 00000101, 00000110, 00000111, 00001001, 00001010, 00001011, 00001100, 00001000, 00001001, 00001010, 00011100, 00011101.
Microprocessors I need EMU8086 - MICROPROCESSOR EMULATOR code to run all given tasks below Task 01Swap two numbers using ADD/SUB instructions only.Task 02If A, B and C are 3 variables, perform the given mathematical operation • A = C + (B – A) - 2 ;Task 03If X, Y and Z are 3 variables, then perform the given mathematical operation• Y * Y / (4 * X * Z)Task 04Perform the following arithmetic operation: 10 * 7 / (1 - 4) + 13 + 52 – 4 *

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